Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e.
$$ t = \sum c_{ij} a_i\otimes b_j$$
for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of operators on a finite-dimensional Hilbert space $H_A$, then apply this property again to $A = H_A\otimes H_A^\ast$ to get a decomposition of such an operator in terms of bras and kets. Additionally, the equation you have written down doesn't make any sense - $\rho^A$ is supposed to be an operator on $A$, but $\langle a_1 \vert a_2 \rangle$ and $\operatorname{tr}(\lvert b_1 \rangle\langle b_2 \rvert)$ seem to be both numbers. You can find the proper definition of the partial trace in the Wikipedia article.
The two different Hamiltonian forms represent different physical things. The first, given by
$$
\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1},
$$
represents the free Hamiltonians of each system separately, and the second, given by
$$
\hat{H}_1\otimes\hat{H}_2\tag{2},
$$
represents an interaction term between two quantum systems. (So, e.g. the first might be the kinetic and potential energies of two individual oscillators, and then the second might represent the interaction Hamiltonian if, for instance, the two oscillators are connected by a spring).
Importantly, Hamiltonians of the form (1) evolve unentangled states into unentangled ones, but Hamiltonians of the form (2) can evolve unentangled states into entangled ones. To see this, consider the following calculations.
Consider two quantum systems described by Hamiltonians $H_1$ and $H_2$, whose eigensystems are separately described by
$$
\hat{H}_1|\psi_n\rangle = \epsilon_n|\psi_n\rangle,~~~~~\hat{H}_2|\phi_n\rangle = \mu_n|\phi\rangle.
$$
Let's consider case 1 first, in which the combined Hamiltonian of the combined system happens to be
\begin{equation}
\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1}
\end{equation}
Then, the eigenstates of this operator are the the product of the eigenstates of the individual Hamiltonians, and the eigenvalues are sums of the individual eigenvalues, which we can show by computing
$$
\left(\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right)
=(\epsilon_n+\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
(The details involve just using linearity.)
Now, given an arbitrary unentangled initial state $|\Psi(0)\rangle$, it can be written as the product of vectors expanded in the individual energy eigenbases as
$$
|\Psi(0)\rangle
=
\left(\sum_{n}a_{n}|\psi_n\rangle\right)\otimes\left(
\sum_{m}b_n|\phi_m\rangle\right)
=
\sum_{nm}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
We can then get the full time-dependence by attaching the exponential factors to the eigenvectors in the usual way, yielding
$$
|\Psi(t)\rangle
=
\sum_{nm}e^{-i(\epsilon_n+\mu_m)t/\hbar}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
Crucially, this state factors as
$$
|\Psi(t)\rangle=\left(\sum_{nm}
e^{-i\epsilon_nt/\hbar}
a_n|\psi_n\rangle\right)\otimes\left(
\sum_{m}
e^{-i\mu_mt/\hbar}
b_m|\phi_m\rangle\right),
$$
and so if the system started unentangled, it remains unentangled. This is purely a consequence of the fact that the Hamiltonian is a sum of single-system Hamiltonians because this is what leads to the eigenenergies being sums of the individual ones, which allows us to factor the exponential.
Now, to see that that doesn't work in the case of a Hamiltonian of the second form, given by
$$
\hat{H}_1\otimes\hat{H}_2\tag{2},
$$
we first note that that the product of eigenvectors is still an eigenvector, but the eigenvalues are now products of the individual eigenvalues, i.e.,
$$
\left(\hat{H}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right)
=(\epsilon_n\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
If we again start with the initially unentangled state shown above, then the full time-dependent state is given by
$$
|\Psi(t)\rangle
=
\sum_{nm}e^{-i(\epsilon_n\mu_m)t/\hbar}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right),
$$
which can no longer be factored in general!
This can also be seen by exponentiating the Hamiltonians directly to get the unitary time-evolution operators. For Hamiltonians of the form (1), the time-evolution operator is
$$
U_1(t) = \exp\left(
-\frac{it}{\hbar}
(\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2)
\right)
=
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1\otimes\hat{I}_2
\right)
\exp\left(
-\frac{it}{\hbar}
\hat{I}_1\otimes\hat{H}_2
\right),
$$
which is allowed because the two operators commute with each other. Furthermore, it is relatively straight-forward to show that this can be written as
$$
U_1(t) =
\left(
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1
\right)
\otimes\hat{I}_2\right)
\left(\hat{I}_1\otimes \exp\left(
-\frac{it}{\hbar}
\hat{H}_2
\right)\right)
=
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1
\right)
\otimes
\exp\left(
-\frac{it}{\hbar}
\hat{H}_2
\right).
$$
Thus, this time-evolution operator can then be seen to "conserve unentangled-ness". The other one does not factor in that same way.
Best Answer
$|\phi(1)\rangle \otimes |\chi(2)\rangle $ is a cumbersome notation to write ket corresponding to $\psi$ function $\phi(\mathbf r_1)\chi(\mathbf r_2)$, where $\mathbf r_i$ refers to coordinates of the $i$-th subsystem. That's why the order of factors in $\otimes$ product does not matter; the resulting ket corresponds to the same $\psi$ function and is thus the same ket.
On the other hand, $|\phi\rangle \otimes |\chi\rangle $ (without labels) is meant to be read according to different convention; here it is commonly understood that the order of factor signifies the sub-system it refers to. So
$|\phi\rangle \otimes |\chi\rangle $ denotes ket corresponding to $\phi(\mathbf r_1)\chi(\mathbf r_2)$ just as $|\phi(1)\rangle \otimes |\chi(2)\rangle $ does, but:
$|\chi\rangle \otimes |\phi\rangle $ denotes ket corresponding to $\chi(\mathbf r_1)\phi(\mathbf r_2)$ which is not the same. This is because different meaning of the $\otimes$ notation is used.